634 CHAPTER 6 The Circular Functions and Their Graphs EXAMPLE 1 Graphing y =sin 1x −d 2 Graph y = sin Ax - p 3 B over one period. SOLUTION Method 1 For the argument x - p 3 to result in all possible values throughout one period, it must take on all values between 0 and 2p, inclusive. To find an interval of one period, we solve the following three-part inequality. 0 … x - p 3 … 2p Three-part inequality p 3 … x … 7p 3 Add p 3 to each part. Use the method described in the previous section to divide the interval C p 3 , 7p 3 D into four equal parts, obtaining the following x-values. p 3 , 5p 6 , 4p 3 , 11p 6 , 7p 3 A table of values using these x-values follows. x p 3 5p 6 4p 3 11p 6 7p 3 x −P 3 0 p 2 p 3p 2 2p sin 1x −P 3 2 0 1 0 -1 0 We join the corresponding points with a smooth curve to obtain the solid blue graph shown in Figure 32. The period is 2p, and the amplitude is 1. These are key x-values. y = sin x 0 1 y = sin x – ( ) x –1 y 3 P p 3 p 6 5p 6 7p 3 4p 3 11p 6 – 2p p The graph can be extended through additional periods by repeating the given portion of the graph, as necessary. Figure 32 Method 2 We can also graph y = sin Ax - p 3 B by using a horizontal translation of the graph of y = sin x. The argument x - p 3 indicates that the graph will be translated to the right p 3 units (the phase shift) compared to the graph of y = sin x. See Figure 32. To graph a function using this method, first graph the basic circular function, and then graph the desired function using the appropriate translation. S Now Try Exercise 39. CAUTION In Example 1, the argument of the function is Ax - p 3 B. The parentheses are important here. If the function had been y = sin x - p 3 , the graph would be that of y = sin x translated vertically down p 3 units.
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